Answer:
[tex]\displaystyle \frac{17}{2}[/tex].
Step-by-step explanation:
Let the [tex]x[/tex]-coordinate of [tex]P[/tex] be [tex]t[/tex]. For [tex]P\![/tex] to be on the graph of the function [tex]y = \sqrt{x}[/tex], the [tex]y[/tex]-coordinate of [tex]\! P[/tex] would need to be [tex]\sqrt{t}[/tex]. Therefore, the coordinate of [tex]P \![/tex] would be [tex]\left(t,\, \sqrt{t}\right)[/tex].
The Euclidean Distance between [tex]\left(t,\, \sqrt{t}\right)[/tex] and [tex](9,\, 0)[/tex] is:
[tex]\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}[/tex].
The goal is to find the a [tex]t[/tex] that minimizes this distance. However, [tex]\sqrt{t^2 - 17 \, t + 81}[/tex] is non-negative for all real [tex]t\![/tex]. Hence, the [tex]\! t[/tex] that minimizes the square of this expression, [tex]\left(t^2 - 17 \, t + 81\right)[/tex], would also minimize [tex]\sqrt{t^2 - 17 \, t + 81}\![/tex].
Differentiate [tex]\left(t^2 - 17 \, t + 81\right)[/tex] with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17[/tex].
[tex]\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2[/tex].
Set the first derivative, [tex](2\, t - 17)[/tex], to [tex]0[/tex] and solve for [tex]t[/tex]:
[tex]2\, t - 17 = 0[/tex].
[tex]\displaystyle t = \frac{17}{2}[/tex].
Notice that the second derivative is greater than [tex]0[/tex] for this [tex]t[/tex]. Hence, [tex]\displaystyle t = \frac{17}{2}[/tex] would indeed minimize [tex]\left(t^2 - 17 \, t + 81\right)[/tex]. This [tex]t\![/tex] value would also minimize [tex]\sqrt{t^2 - 17 \, t + 81}\![/tex], the distance between [tex]P[/tex] [tex]\left(t,\, \sqrt{t}\right)[/tex] and [tex](9,\, 0)[/tex].
Therefore, the point [tex]P[/tex] would be closest to [tex](9,\, 0)[/tex] when the [tex]x[/tex]-coordinate of [tex]P\![/tex] is [tex]\displaystyle \frac{17}{2}[/tex].
